A space form is defined as a manifold admitting a Riemannian manifold of constant sectional curvature

A classical result of Cartan states that a manifold is a space form if and only if it is a quotient of $S^n$, $\mathbb{R}^n$, or $\mathbb{H}^n$ with their usual metrics by a discrete group of isometries $\Gamma$ acting properly discontinuously; furthermore $\Gamma$ is isomorphic to the fundamental group of the space form. This reduces the space form problem to a problem in group theory.

Such a discrete group in the case of $\mathbb{R}^n$ or $\mathbb{H}^n$ are referred to respectively as Kleinian or crystallographic.

The sphere and the real projective space are the only even-dimensional spherical space forms.

I have often seen papers in differential geometry, for example convergence results for Ricci flow [cf. Hamilton 1982 & 1986, Böhm-Wilking, Brendle-Schoen], show that manifolds admitting metrics of a certain type (e.g. of positive Ricci curvature, positive curvature operator, positive isotropic curvature) are space forms, "which have been completely classified" [cf. Wolf's book Spaces of constant curvature], seeming to imply that this gives a complete understanding of such manifolds.

However, even having seen such a list of groups, I don't feel that it gives me a much better understanding of space forms than what I listed above. How can I understand it better? Some sample questions: (I'm happy to restrict to the three-dimensional case for simplicity.)

Are two space forms diffeomorphic if their fundamental groups are isomorphic? If this is not true, what if we require more of their homotopy groups to be isomorphic?

Maybe a natural class of 3-manifolds to consider are those with locally homogeneous metrics (i.e. for any $p,q\in M$ there are neighborhoods $U_p\ni p$ and $V_q\ni q$ and an isometry $U_p\to V_q$); by a result of Singer [CPAM 1960] the universal cover of such a manifold with the pulled back metric is homogeneous and so is one of the eight geometries. If this geometry is $S^2\times\mathbb{R}$, $\mathbb{H}^2\times\mathbb{R}$ or $\widetilde{\operatorname{SL}}(2,\mathbb{R})$ then it is immediate from the second bullet point above that $M$ is not a space form; however I think (?) Nil and Sol have the topology of $\mathbb{R}^3$; so is it possible for a locally homogeneous manifold of one of these types to be a space form?

More generally, given any closed 3-manifold, are there any generally applicable (but not too trivial) necessary or sufficient conditions to see if it is a space form, perhaps on the level of homotopy, homology, or cohomology?

Given any compact 3-manifold $M$, when can we make $M\sharp N$ a space form for some choice of compact 3-manifold $N$?

These are just a handful of things I'm curious about. My main question is the title.

1 Answer
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Are two space forms diffeomorphic if their fundamental groups are
isomorphic? If this is not true, what if we require more of their
homotopy groups to be isomorphic?

This is true in the hyperbolic case. In fact more is true: in dimensions three and higher Mostow rigidity implies that if $M$ and $N$ have isomorphic fundamental groups then they are isometric.

It is false in the spherical case. There are non-homeomorphic lens spaces with isomorphic (finite, cyclic) fundamental groups.

Nil and Sol have the topology of $\mathbb{R}^3$; so is it possible for a locally
homogeneous manifold of one of these types to be a space form?

No. If a closed three-manifold is geometric, then it has a unique geometry. (See the next answer for why.)

More generally, given any closed 3-manifold, are there any generally
applicable (but not too trivial) necessary or sufficient conditions to
see if it is a space form, perhaps on the level of homotopy, homology,
or cohomology?

The class of geometric three-manifolds can be characterized topologically (mostly in terms of their JSJ decomposition) or algebraically. Furthermore, each of the eight geometries has a topological, and also algebraic, description within this framework. For example, a closed three-manifold is a spherical space form if and only if it has finite fundamental group. A closed three-manifold is a Euclidean space form if and only if its fundamental group has $\mathbb{Z}^3$ as a finite index subgroup.

Given any compact 3-manifold $M$, when can we make $M \# N$ a space form for
some choice of compact 3-manifold $N$?

The only closed geometric three-manifold that is a connect sum is $\mathbb{RP}^3 \# \mathbb{RP}^3$. It has $S^2 \times \mathbb{R}$ geometry.

An over-all answer to your question might be: The hyperbolic manifolds are hugely interesting to a range of mathematicians, and are still actively being studied.

The simple argument for the given case is as follows: in the NIL and SOL cases, the fundamental group is virtually solvable and not virtually abelian; for space forms: in the hyperbolic case it's not virtually solvable, and in the Euclidean or spherical case it's virtually abelian. A more refined argument, which also separates these cases (NIL/SOL vs constant curvature) up to quasi-isometry, is that for space forms the Dehn function is at most quadratic while in NIL and SOL case it's at least cubic.
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YCorMar 23 '14 at 11:51

@SamNead Thank you for your answer! I'm confused by one point: if we take for granted "a closed three-manifold is a spherical space form if and only if it has finite fundamental group" then wouldn't the Poincaré conjecture be immediate from the classification of spherical space forms? Or is this a consequence of the geometrization theorem?
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youlerMar 23 '14 at 20:01

@YvesCornulier Your answer is a bit over my head, could you expand it to a full answer? For example where do these fundamental group restrictions come from?
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youlerMar 23 '14 at 20:02